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a: \(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{3}{4}=3-\dfrac{1}{2}x-\dfrac{1}{2}-\dfrac{1}{3}x-\dfrac{2}{3}\)
=>13/12x=13/12
hay x=1
b: \(\Leftrightarrow\dfrac{3x-11}{11}-\dfrac{x}{3}=\dfrac{3x-5}{7}-\dfrac{5x-3}{9}\)
\(\Leftrightarrow\dfrac{3}{11}x-1-\dfrac{1}{3}x=\dfrac{3}{7}x-\dfrac{5}{7}-\dfrac{5}{9}x+\dfrac{1}{3}\)
\(\Leftrightarrow x\cdot\dfrac{46}{693}=\dfrac{13}{21}\)
hay x=429/46
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x^11+x^4+1
=x^11-x^2+x^4-x+x^2+x+1
=x^2(x^9-1)+x(x^3-1)+(x^2+x+1)
=x^2[(x^3-1)(x^6+x^3+1)]+x(x-1)(x^2+x+1)+(x^2+x+1)
=x^2(x-1)(x^2+x+1)(x^6+x^3+1)+x(x-1)(x^2+x+1)+(x^2+x+1)
=(x^2+x+1)[x^2(x-1)(x^6+x^3+1)+x(x-1)+1]
=(x^2+x+1)(x^9-x^8+x^6-x^5+x^3-x+1)
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a)
\(\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\frac{5}{2}\)
\(\Leftrightarrow3x^3-x^2+x-3x^2+x-1+4x^2-3x^3=\frac{5}{2}\)
\(\Leftrightarrow2x-1=\frac{5}{2}\Leftrightarrow2x=1+\frac{5}{2}=\frac{7}{2}\Leftrightarrow x=\frac{7}{4}\)
b)
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
\(\Leftrightarrow4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)
\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
\(\Leftrightarrow8x+4-4x+1+8=11\Leftrightarrow4x+13=11\Leftrightarrow4x=-2\Leftrightarrow x=-\frac{1}{2}\)
c)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow-4x+1+6x+9+245=0\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{255}{2}\).
a ) ( 3x2 - x + 1 ) ( x + 1 ) + x2 ( 4 - 3x ) = 5/2
=> 3x3 + 3x2 - x2 - x + x + 1 + 4x2 - 3x3 = 5/2
=> 6x2 + 1 = 5/2
=> 6x2 = 1,5
=> x2 = 0,25
=> x = 0,5
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a, (x-1)^2 + (x+3)^2 = 2(x-2)(x+1) + 38
<=> x^2 -2x +1 + x^2 + 6x +9 = 2x^2 +2x -4x -4 +38
<=> x^2 -2x +x^2 +6x -2x^2 -2x +4x= -4 +38 -10
<=> 6x= 24
<=> x = 4
=> S={4}
b, 5(2x-3)-4(5x-7)= 19 -2(x+11)
<=> 10x -15 -20x +28 = 19-2x-22
<=> 10x -20x +2x = 19 -22 +15 -28
<=> -8x = -16
<=> x = 2
=> S={2}
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\(x^{11}+x^7+1\)
\(=\left(x^{11}-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^9-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^3-1\right)\left(x^6+x^3+1\right)+x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^6+x^3+1\right)+x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)\left(x^6+x^3+1\right)+x\left(x-1\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left(x^9+x^6+x^3-x^8-x^5-x^2+x^5+x^2-x^4-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^9-x^8+x^6-x^4+x^3-x+1\right)\)
\(x^7+x^2+1\)
\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x^4+x\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^5+x^2-x^4-x+1\right)\)
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a) \(\left(x-11\right)+\frac{3x}{x-11}=3+\frac{33}{x-11}\)
\(\Leftrightarrow x+\frac{3x}{x-11}-\frac{33}{x-11}=14\)
\(\Leftrightarrow x^2-11x+3x-33=14x-154\)
\(\Leftrightarrow x^2-22x+121=0\)
\(\Leftrightarrow\left(x-11\right)^2=0\Leftrightarrow x=11\)
Vậy .......
b) \(\frac{7-2x}{x-1}=\frac{1-4x}{x+2}\Leftrightarrow\left(7-2x\right)\left(x+2\right)=\left(1-4x\right)\left(x-1\right)\)
\(\Leftrightarrow7x-2x^2+14-4x=x-4x^2-1+4x\)
\(\Leftrightarrow2x^2=-15\)(vô lí)
Vậy pt vô nghiệm
c) \(\frac{3-2x}{x+1}=2+\frac{1-4x}{x-2}\)
\(\Leftrightarrow\left(3-2x\right)\left(x-2\right)=2\left(x+1\right)\left(x-2\right)+\left(1-4x\right)\left(x+1\right)\)
\(\Leftrightarrow3x-2x^2-6x+4x=2x^2+2x-4x-4+x-4x^2+1-4x\)
\(\Leftrightarrow6x=-3\Leftrightarrow x=-\frac{1}{2}\)
Vậy.........
(gửi trước 3 câu)
d) \(\frac{109x-4}{111x+1}-1=0\Leftrightarrow109x-4=111x+1\Leftrightarrow2x=-5\Leftrightarrow x=-\frac{5}{2}\)
Vậy x=-5/2
e) \(\frac{x^2-7}{x}=x-\frac{1}{2}\Leftrightarrow\frac{x^2-7}{x}-\frac{x^2}{x}=-\frac{1}{2}\Leftrightarrow-\frac{7}{x}=\frac{1}{2}\Leftrightarrow x=-14\)
f) \(\frac{x+1}{x+2}=3\Leftrightarrow x+1=3x+6\Leftrightarrow2x=7\Leftrightarrow x=\frac{7}{2}\)
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a: \(\dfrac{-7}{x^2-4}=\dfrac{-7}{\left(x-2\right)\left(x+2\right)}=\dfrac{-14}{2\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{11}{2x+4}=\dfrac{11}{2\left(x+2\right)}=\dfrac{11\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}\)
b: \(\dfrac{2}{9x^2-1}=\dfrac{2}{\left(3x-1\right)\left(3x+1\right)}\)
\(\dfrac{4x}{1-3x}=\dfrac{-4x}{3x-1}=\dfrac{-4x\left(3x+1\right)}{\left(3x-1\right)\left(3x+1\right)}\)
c: \(\dfrac{3}{x+2}=\dfrac{6\left(x^2-2x+4\right)}{2\left(x+2\right)\left(x^2-2x+4\right)}\)
\(\dfrac{x+1}{x^3+8}=\dfrac{2x+2}{2\left(x+1\right)\left(x^2-2x+4\right)}\)
\(\dfrac{x+2}{2\left(x+2\right)}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{2\left(x+2\right)\left(x^2-2x+4\right)}\)
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Bài 1:
\(a,3\left(x-11\right)-2\left(x+11\right)=2011\)
\(\Leftrightarrow3x-33-2x-22=2011\)
\(\Leftrightarrow x-55=2011\)
\(\Leftrightarrow x=2066\)
Vậy pt có nghiệm x = 2066
\(b,\left(x-1\right)\left(3x-7\right)=\left(x-1\right)\left(x+30\right)\)
\(\Leftrightarrow\left(x-1\right)\left(3x-7\right)-\left(x-1\right)\left(x+30\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-7-x-30\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-37\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-37=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{37}{2}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{1;\dfrac{37}{2}\right\}\)
\(c,\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\) (1)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{0;-1\right\}\)
\(d,\left|2x-3\right|=x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x+1\\2x-3=-x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=1+3\\2x+x=-1+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{4;\dfrac{2}{3}\right\}\)
Bài 2:
\(a,2\left(x-1\right)< x+1\)
\(\Leftrightarrow2x-2< x+1\)
\(\Leftrightarrow2x-x< 1+2\)
\(\Leftrightarrow x< 3\)
Vậy bpt có nghiệm x < 3
b, Đề bài ko rõ
x-\(\dfrac{x+2}{3}\)nhỏ hơn hoặc bằng 3x+\(\dfrac{x}{2}+5\)
đề yêu cầu j